问题
I am new to kdb and researching it for a use case to generate time series data using a table of various function inputs. Each row of the table consists of function inputs keyed by an id and segment and will call one function per row. I have figured out how to identify which function albeit using brute force nested conditions.
My question is 2 part
- How does one employ kicking off the execution of these functions?
- Once the time series data is generated for each id and segment, how best can the the output be compiled into a singular table (sample output noted below - I have thought about one table for each id and then compile in two steps which would work as well but we'll have thousands of ids)
Below is a sample table and some conditions to add meta data including which function to apply
//Create sample table and add columns to identify unknown and desired function
t:([id:`AAA`AAA`AAA`BBB`CCC;seg:1 2 3 1 1];aa: 1500 0n 400 40 900;bb:0n 200 30 40 0n;cc: .40 .25 0n 0n .35)
t: update Uknown:?[0N = aa;`aa;?[0N = bb;`bb;?[0N = cc;`cc;`UNK]]] from t
t: update Call_Function:?[0N = aa;`Solveaa;?[0N = bb;`Solvebb;?[0N = cc;`Solvecc;`NoFunction]]] from t
A sample function below uses the inputs from table t to generate time series data (limited to 5 periods for example here) and test using X
//dummy function to generate output for first 5 time periods
Solvebb:{[aa;cc]
(aa%cc)*(1-exp(neg cc*1+til 5))
}
//test the function as an example for dummy output in result table below
x: flip enlist Solvebb[1500;.40] //sample output for AAA seg1 from t for example
The result would ideally be a sample table similar to below
t2: `id`seg xkey ("SIIIS";enlist",") 0:`:./Data/sampleOutput.csv
id seg| seg_idx tot_idx result
-------| ------------------------
AAA 1 | 1 1 1,236.30
AAA 1 | 2 2 2,065.02
AAA 1 | 3 3 2,620.52
AAA 1 | 4 4 2,992.89
AAA 1 | 5 5 3,242.49
AAA 2 | 1 6
AAA 2 | 2 7
AAA 2 | 3 8
AAA 2 | 4 9
AAA 2 | 5 10
AAA 3 | 1 11
AAA 3 | 2 12
AAA 3 | 3 13
AAA 3 | 4 14
AAA 3 | 5 15
BBB 1 | 1 1
BBB 1 | 2 2
BBB 1 | 3 3
BBB 1 | 4 4
BBB 1 | 5 5
..
回答1:
It's difficult without more details, but something like the following may help.
First, it may be easier to define Solvebb so that it can take 3 inputs and simple ignores the middle one:
q)Solvebb:{[aa;bb;cc](aa%cc)*(1-exp(neg cc*1+til 5))}
And adding dummy functions for the other two in your table (NB. it's important for the use of ungroup later that the output of these functions are lists):
q)Solveaa:{[aa;bb;cc] (bb+cc;bb*cc)}
q)Solvecc:{[aa;bb;cc] (aa+bb;aa*bb)}
You can apply each call each function on all three vectors of input with:
q)update result:first[Call_Function]'[aa;bb;cc] by Call_Function from t
id seg| aa bb cc Uknown Call_Function result
-------| -------------------------------------------------------------------------------
AAA 1 | 1500 0.4 bb Solvebb 1236.3 2065.016 2620.522 2992.888 3242.493
AAA 2 | 200 0.25 aa Solveaa 200.25 50
AAA 3 | 400 30 cc Solvecc 430 12000f
BBB 1 | 40 40 cc Solvecc 80 1600f
CCC 1 | 900 0.35 bb Solvebb 759.3735 1294.495 1671.589 1937.322 2124.581
and you can unravel this table by applying the ungroup function
q)ungroup update result:first[Call_Function]'[aa;bb;cc] by Call_Function from t
id seg aa bb cc Uknown Call_Function result
---------------------------------------------------
AAA 1 1500 0.4 bb Solvebb 1236.3
AAA 1 1500 0.4 bb Solvebb 2065.016
AAA 1 1500 0.4 bb Solvebb 2620.522
AAA 1 1500 0.4 bb Solvebb 2992.888
AAA 1 1500 0.4 bb Solvebb 3242.493
AAA 2 200 0.25 aa Solveaa 200.25
AAA 2 200 0.25 aa Solveaa 50
AAA 3 400 30 cc Solvecc 430
AAA 3 400 30 cc Solvecc 12000
BBB 1 40 40 cc Solvecc 80
BBB 1 40 40 cc Solvecc 1600
CCC 1 900 0.35 bb Solvebb 759.3735
CCC 1 900 0.35 bb Solvebb 1294.495
CCC 1 900 0.35 bb Solvebb 1671.589
CCC 1 900 0.35 bb Solvebb 1937.322
CCC 1 900 0.35 bb Solvebb 2124.581
来源：https://stackoverflow.com/questions/63163687/kdb-referencing-functions-from-a-table